Question

$I(x) = \int \frac{x^2(xsec^2x + tanx)}{(xtanx+1)^2} dx$. If $I(0)=1$ then $I(\frac{\pi}{4})$ is equal to

• A. $-\frac{\pi^2}{4\pi+16}+2ln(\frac{\pi+4}{4\sqrt2})+1$
• B. $\frac{\pi^2}{4\pi+16}-2ln(\frac{\pi+4}{4\sqrt2})+1$
• C. $-\frac{\pi^2}{\pi+4}+2ln(\frac{\pi+1}{\sqrt2})+1$
• D. $\frac{\pi^2}{\pi+16}+2ln(\frac{\pi+1}{4\sqrt2})+1$

Answer 1

Answer: (A)

$-\frac{\pi^2}{4\pi+16}+2ln(\frac{\pi+4}{4\sqrt2})+1$

Answer 2

Using integration by parts
$I(x) = x^2.\frac{(-1)}{x\:tanx+1}\:-\:\int\:2x.\frac{(-1)}{x\:tanx+1}dx$
$=\,-\frac{x^2}{x\:tanx+1}+2\int\frac{x\:cos\,x}{x\:sin\,x+cos\,x}dx$
$\Rightarrow I(x)=-\frac{x^2}{x\:tan\,x+1}+2ln|x\:sin\,x+cos\,x|+c$
put $x=0$
$c=1$
$\therefore I(\frac{\pi}{4})=\frac{\frac{-\pi^2}{16}}{\frac{\pi}{4}+1}+2ln(\frac{\frac{\pi}{4}+1}{\sqrt{2}})+1$
$I(\frac{\pi}{4})= -\frac{\pi^2}{4\pi+16}+2ln(\frac{\pi+4}{4\sqrt{2}})+1$

So, the correct answer is (A): $-\frac{\pi^2}{4\pi+16}+2ln(\frac{\pi+4}{4\sqrt2})+1$